Date difference with leap year
I have 5 different schedules for 5 weeks:
- first week = Monday to Friday (8am to 5pm) && Rest days on Saturday and Sunday
- second week = Monday to Friday (10am to 6pm) && Rest days on Saturday and Sunday
- third week = Monday to Friday (11am to 7pm) && Rest days on Saturday and Sunday
- fourth week 开发者_StackOverflow中文版= Monday Rest Day && Tuesday to Saturday (10:30 am to 6:30pm) && Sunday Rest Day
- fifth week = Monday Rest Day && Tuesday to Saturday (8:30 am to 5:30pm) && Sunday Rest Day
Base on my calculation array [0],[0] which is Monday of first week is set to April 25, 2011.
I have this code to compute the difference between input date and start date, which is April 25, 2011.
$tdays = floor((strtotime($date2) - strtotime($date1))/86400);
I could now compute my work schedule starting April of 2011 up until February of 2012. However if I enter a date beyond February 2012, the output is wrong due to leap year. Is there a technique for this?
If you are able to make use of php 5.3 you should use date_diff()
or try something like this :
<?php
function dateDifference($startDate, $endDate)
{
$startDate = strtotime($startDate);
$endDate = strtotime($endDate);
if ($startDate === false || $startDate < 0 || $endDate === false || $endDate < 0 || $startDate > $endDate)
return false;
$years = date('Y', $endDate) - date('Y', $startDate);
$endMonth = date('m', $endDate);
$startMonth = date('m', $startDate);
// Calculate months
$months = $endMonth - $startMonth;
if ($months <= 0) {
$months += 12;
$years--;
}
if ($years < 0)
return false;
// Calculate the days
$offsets = array();
if ($years > 0)
$offsets[] = $years . (($years == 1) ? ' year' : ' years');
if ($months > 0)
$offsets[] = $months . (($months == 1) ? ' month' : ' months');
$offsets = count($offsets) > 0 ? '+' . implode(' ', $offsets) : 'now';
$days = $endDate - strtotime($offsets, $startDate);
$days = date('z', $days);
return array($years, $months, $days);
}
?>
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